In case $AB\parallel CD$, symmetry wrt perpendicular bisector of $AB$ solves the problem. Now we will assume $AB$ and $CD$ are not parallel. Let $Q=AB\cap CD$. By simple angle chasing, $QX=QY$ or $Q$ lies on the perpendicular bisector of $XY$. Since $MX=MY$, $M$ also lies on this perpendicular bisector. Note that $BC$ and $AD$ are antiparallel wrt $\angle AQD$. This means that the line $QM$, which is the median of $\triangle QAB$ is the symmedian of triangle $QBC$. In other words, lines $QM$ and $QN$ are symmetric wrt bisector of line $AQD$. If we look at it at another angle, $QM$ is the angle bisector of $\angle XQY$ (due to symmetry wrt perpendicular bisector of $XY$). So reflection of line $QM$ wrt angle bisector $\angle XQY$ should be $QM$ itself. But earlier, we found it is $QN$. Thus, line $QM$ and $QN$ coincide or $Q,M,N$ are collinear. Since $Q$ and $M$ lie on the perpendicular bisector of $XY$, $N$ also does. So $NX=NY$

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